√3 cot_2theta+4 cottheta+√3=0 এর সমাধান কোনটি?

Explanation:

Another Explanation (5): সমাধান: \[ \sqrt{3} \cot 2\theta + 4 \cot \theta + \sqrt{3} = 0 \] আমরা জানি, \(\cot 2\theta = \frac{\cot^2 \theta - 1}{2 \cot \theta}\) সুতরাং, \[ \sqrt{3} \frac{\cot^2 \theta - 1}{2 \cot \theta} + 4 \cot \theta + \sqrt{3} = 0 \] ধরি, \(x = \cot \theta\). তাহলে, \[ \sqrt{3} \frac{x^2 - 1}{2x} + 4x + \sqrt{3} = 0 \] \[ \sqrt{3} (x^2 - 1) + 8x^2 + 2\sqrt{3}x = 0 \] \[ \sqrt{3}x^2 - \sqrt{3} + 8x^2 + 2\sqrt{3}x = 0 \] \[ (8 + \sqrt{3})x^2 + 2\sqrt{3}x - \sqrt{3} = 0 \] এটি একটি দ্বিঘাত সমীকরণ। সুতরাং, \[ x = \frac{-2\sqrt{3} \pm \sqrt{(2\sqrt{3})^2 - 4(8+\sqrt{3})(-\sqrt{3})}}{2(8+\sqrt{3})} \] \[ x = \frac{-2\sqrt{3} \pm \sqrt{12 + 32\sqrt{3} + 12}}{16+2\sqrt{3}} \] \[ x = \frac{-2\sqrt{3} \pm \sqrt{24 + 32\sqrt{3}}}{16+2\sqrt{3}} \] \[ x = \frac{-2\sqrt{3} \pm 2\sqrt{6 + 8\sqrt{3}}}{16+2\sqrt{3}} \] এই পর্যন্ত ক্যালকুলেশনটি জটিল হয়ে যাচ্ছে। অন্যভাবে চেষ্টা করি। \[ \sqrt{3} \cot 2\theta + 4 \cot \theta + \sqrt{3} = 0 \] \[ \sqrt{3} \cot 2\theta + \sqrt{3} + 4 \cot \theta = 0 \] \[ \sqrt{3} (\cot 2\theta + 1) + 4 \cot \theta = 0 \] \[ \sqrt{3} \left( \frac{\cos 2\theta}{\sin 2\theta} + 1 \right) + 4 \frac{\cos \theta}{\sin \theta} = 0 \] \[ \sqrt{3} \left( \frac{\cos 2\theta + \sin 2\theta}{\sin 2\theta} \right) + 4 \frac{\cos \theta}{\sin \theta} = 0 \] \[ \sqrt{3} \left( \frac{\cos 2\theta + \sin 2\theta}{2 \sin \theta \cos \theta} \right) + 4 \frac{\cos \theta}{\sin \theta} = 0 \] \[ \frac{\sqrt{3}}{2} \left( \frac{\cos 2\theta + \sin 2\theta}{\sin \theta \cos \theta} \right) + 4 \frac{\cos \theta}{\sin \theta} = 0 \] \[ \frac{\cos \theta}{\sin \theta} \left[ \frac{\sqrt{3}}{2} \left( \frac{\cos 2\theta + \sin 2\theta}{\cos^2 \theta} \right) + 4 \right] = 0 \] যদি \(\cot \theta = 0\) হয়, \(\theta = \frac{\pi}{2} + n\pi\) নতুবা, \[ \frac{\sqrt{3}}{2} \left( \frac{\cos 2\theta + \sin 2\theta}{\cos^2 \theta} \right) + 4 = 0 \] \[ \sqrt{3} (\cos 2\theta + \sin 2\theta) + 8 \cos^2 \theta = 0 \] \[ \sqrt{3} (\cos 2\theta + \sin 2\theta) + 4(1 + \cos 2\theta) = 0 \] \[ \sqrt{3} \cos 2\theta + \sqrt{3} \sin 2\theta + 4 + 4 \cos 2\theta = 0 \] \[ (4 + \sqrt{3}) \cos 2\theta + \sqrt{3} \sin 2\theta = -4 \] ধরি, \(R \cos (2\theta - \alpha) = (4 + \sqrt{3}) \cos 2\theta + \sqrt{3} \sin 2\theta \) যেখানে, \(R = \sqrt{(4+\sqrt{3})^2 + (\sqrt{3})^2} = \sqrt{16 + 8\sqrt{3} + 3 + 3} = \sqrt{22 + 8\sqrt{3}}\) এবং \(\tan \alpha = \frac{\sqrt{3}}{4+\sqrt{3}}\) তাহলে, \(R \cos (2\theta - \alpha) = -4\) \[ \cos (2\theta - \alpha) = \frac{-4}{R} = \frac{-4}{\sqrt{22 + 8\sqrt{3}}} \) এখন, অপশনটি যাচাই করি: \(\theta = n\pi + \frac{2\pi}{3}\) যদি \(n = 0\), \(\theta = \frac{2\pi}{3}\) \[ \sqrt{3} \cot \frac{4\pi}{3} + 4 \cot \frac{2\pi}{3} + \sqrt{3} = \sqrt{3} \cdot \frac{1}{\sqrt{3}} + 4 \cdot \frac{-1}{\sqrt{3}} + \sqrt{3} = 1 - \frac{4}{\sqrt{3}} + \sqrt{3} = 1 - \frac{4}{\sqrt{3}} + \frac{3}{\sqrt{3}} = 1 - \frac{1}{\sqrt{3}} \ne 0 \] তাহলে প্রদত্ত উত্তরটি সঠিক নয়।